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I have a question, suppose that $S_{n+2}=13S_{n+1}+48S_n$. How do I find a general solution for this recurrence equation and how do I find the particular solution where $S_0=1$ and $S_1=5$.

Here is what I've got so far, I brought everything to the left side of the equation to get $x^2-13x-48=0$. Roots are $16$ and $-3$. So to get the general equation, I believe I have to use the roots somewhere such as $S_n=A\cdot16^n-B\cdot3^n$. I know I am probably wrong.

Could someone explain the process how to get the general solution and particular solution?

JRN
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Rich Sanchez
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1 Answers1

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The general solution to the homogeneous recurrence is $S_n=A\cdot16^n+B(-3)^n$. You can’t pull the sign out of $(-3)^n$: it’s equal to $(-1)^n\cdot3^n$, not to $-3^n$. No matter what the initial values are, the general term will have the form $S_n=A\cdot16^n+B(-3)^n$: each pair of values for $S_0$ and $S_1$ determines a unique pair of coefficients $A$ and $B$, and vice versa.

To find $A$ and $B$, substitute $n=0$ and $n=1$ into the general solution and use the known values of $S_0$ and $S_1$ to get the system

$$\left\{\begin{align*} &1=A+B\\ &5=16A-3B\;. \end{align*}\right.$$

To complete the solution, just solve the system to get $A$ and $B$.

Brian M. Scott
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  • Ok, just for layman's terms. Is the formula to find the solution basically S_n = A(r)^n + B(r)^n where A and B are constants and r is the root of the equation? Or am I wrong? – Rich Sanchez Nov 04 '13 at 04:10
  • @Rich: If the characteristic equation has distinct roots $r_1$ and $r_2$, the general solution is $S_n=Ar_1^n+Br_2^n$. If the characteristic equation has a double root $r$ (e.g., $x^2-4x-4=0$, with double root $2$), the general solution is $S_n=Ar^n+Bnr^n=(A+Bn)r^n$. – Brian M. Scott Nov 04 '13 at 04:14
  • Wonderful! You make it so easy to learn... LOL Another question, how do you know when you can't use a root? I remember him writing it in class but i was behind on the notes and still writing from earlier so didn't get to understand it or really see the example. Is there a case when a characteristic equation is factorable but you can't use one of the roots? – Rich Sanchez Nov 04 '13 at 07:24
  • @Rich: You can always use the roots. Is it possible that he was talking about the case of a repeated root $r$ and saying that you can’t write the general solution as $Ar^n+Br^n$, but have to use $Ar^n+Bnr^n$ instead? – Brian M. Scott Nov 04 '13 at 07:26
  • Not sure, he went fast... LOL – Rich Sanchez Nov 04 '13 at 07:28
  • @Rich: That’s the only thing I can think of that might fit. As long as you’re looking at a homogeneous linear recurrence with constant coefficients, you’ll use all of the roots of the characteristic equation. Of course some of them might not be real numbers: you might get complex roots. – Brian M. Scott Nov 04 '13 at 07:30
  • Ok while I'm on a role understanding, can you tell me how from another post, discrete math What formula did he use to get that general equation? Not the one you just used for mine, right? – Rich Sanchez Nov 04 '13 at 07:38
  • @Rich: That recurrence is non-homogeneous, so it does require different techniques. It can be solved in a variety of ways, and I can’t guess which one(s) you’ll be shown. – Brian M. Scott Nov 04 '13 at 07:41
  • It's not in the Ar^n + Br_2^n model though? Share the simplest layman's terms to find the general equation for that one. – Rich Sanchez Nov 04 '13 at 07:43
  • @Rich: Take a look at the method used in the example in this answer. It could be converted to a general formula for this very restricted type of recurrence, but it’s easier just to remember the method. – Brian M. Scott Nov 04 '13 at 07:49
  • Ok thanks. Sorry forgot about this problem. So how do I get a particular solution from here? – Rich Sanchez Nov 04 '13 at 07:54
  • @Rick: I’m not sure what you mean here by particular equation. – Brian M. Scott Nov 04 '13 at 07:58
  • Sorry meant particular solution... – Rich Sanchez Nov 04 '13 at 08:00
  • @Rich: The general solution is $S_n=A\cdot16^n+B(-3)^n$, and you’ve now found $A$ and $B$, so all you have to do is plug them in to the general solution. However, you need to check your work in your solution of the system: you got $A$ upside down and then calculated $B$ from that incorrect value of $A$. – Brian M. Scott Nov 04 '13 at 08:08
  • OK.. LOL thanks... will work on it now and show if what I got... – Rich Sanchez Nov 04 '13 at 08:08
  • S_n = (8/19)16^n + 11/19(-3)^n? – Rich Sanchez Nov 04 '13 at 08:14
  • @Rich: Yes, that looks much better. – Brian M. Scott Nov 04 '13 at 08:15
  • Ok, thank you much! I can sleep well tonight... LOL – Rich Sanchez Nov 04 '13 at 08:15
  • @Rich: You’re welcome! Get some sleep. :-) – Brian M. Scott Nov 04 '13 at 08:16